Anonymous0 31

Hello I have a question.
Can I set up a strategy for playing roulette where I bet on red and black at the same time in differing values depending on the last 9 rolls?

 trizeroyouroul.com80 704

Hello.
Generally yes. Depending on the special needs.

The function optimize chances calculates the differential bet, if you mean this.
The roulette strategy Differential d’Alembert works with this concept.

 Paroli2 7

Anonymous:

... depending on the last 9 rolls?

This sounds like a tendency concept right? Why exactly 9 rolls?

 trizeroyouroul.com80 704

Paroli:

Why exactly 9 rolls?

Most likely you won't get answer, because that question came from anonymous.

I would say, people don't await some streaks like 9 times even or black.
This is a misunderstanding, because you can't know when very unlikely constellations show up...

I realize the OP won't be back, but this caught my eye.
What would be the most beneficial ratio(s) of wagers from red to black (or vice versa) using this strategy?

 Paroli2 7

First of all it is never a good idea to bet opposite chances at the same time.
Don't bet red and black, odd and even and all corresponding other combinations.

Usually the progression is more important than the chance.
On the long run, you need to have a lot of money, to cover loss streaks of your progression.

Thanks, Paroli. Never tried this strategy before, and will stay away from it in the future too.
(Sometimes I will do black/red, etc. with individual numbers from the opposite group - but usually, I'm a fairly consistent 50/50 martingale kind of guy).
 Paroli2 7

Paroli:

Don't bet red and black, odd and even and so on.

If you do that, you bet 36 (2x 18) numbers out of 37.
All time you bet 1 unit on red and 1 unit on black, you get 2 units back all the time, as long as Zero is not coming. You don't win anything.
When Zero comes, then your 2 units are lost (depending on zero handling).

 cabyxiaomi0 1

When it comes to betting on red or black, choosing both is not possible because you'll end up betting on all the numbers on the table.